# physics question help on work done

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#1

8 Dan has a mass of 62 kg. He is travelling at a speed of 11 m s–1 at the top of a
rollercoaster loop, which is 21.8 m above the ground.
a Calculate:
i his kinetic energy at the top of the loop (1 mark)
ii his change in gravitational potential energy between the top and bottom
of the loop 0.8 m above the ground (2 marks)
iii his kinetic energy at the bottom of the loop, stating any assumption you
iv his speed at the bottom of the loop. (2 marks)
b Dan started the ride 5 m above the ground.
i Calculate the minimum work done by the accelerating force to accelerate
him to the top of the loop in part a. (2 marks)
ii Explain why the work done will be more than this minimum
0
5 years ago
#2
(Original post by lovelyqueen)

8 Dan has a mass of 62 kg. He is travelling at a speed of 11 m s–1 at the top of a
rollercoaster loop, which is 21.8 m above the ground.
a Calculate:
i his kinetic energy at the top of the loop (1 mark)
ii his change in gravitational potential energy between the top and bottom
of the loop 0.8 m above the ground (2 marks)
iii his kinetic energy at the bottom of the loop, stating any assumption you
iv his speed at the bottom of the loop. (2 marks)
b Dan started the ride 5 m above the ground.
i Calculate the minimum work done by the accelerating force to accelerate
him to the top of the loop in part a. (2 marks)
ii Explain why the work done will be more than this minimum
What are your thoughts? What have you done so far?
0
5 years ago
#3
(Original post by lovelyqueen)

8 Dan has a mass of 62 kg. He is travelling at a speed of 11 m s–1 at the top of a
rollercoaster loop, which is 21.8 m above the ground.
a Calculate:
i his kinetic energy at the top of the loop (1 mark)
ii his change in gravitational potential energy between the top and bottom
of the loop 0.8 m above the ground (2 marks)
iii his kinetic energy at the bottom of the loop, stating any assumption you
iv his speed at the bottom of the loop. (2 marks)
b Dan started the ride 5 m above the ground.
i Calculate the minimum work done by the accelerating force to accelerate
him to the top of the loop in part a. (2 marks)
ii Explain why the work done will be more than this minimum
a)
i. kinetic energy is 1/2mv2 = 1/2 x 62 x 112 = 3751 Joules
ii. GPE at top of loop = mass x vertical height = 62 x 21.8 = 1351.6 joules
GPE at the bottom of loop = 62 x 0.8 = 49.6 joules
iii. 1/2 x 62 x 112 = 3751 joules, assuming he isn't accelerating/ is travelling at the same speed.
iv. kinetic energy = 1/2 x 62 x velocity2
(3751 x 2)/62 = velocity2
velocity=11m/s
b)
i. change in energy = work done
energy at 5m = 62 x 5 = 310 joules
energy at 21.8m = 1351.6
change = 1041.6
work done = 1041.6
ii.there will be friction to overcome.

everything should be correct but there could be a mistake in a)iii, iv and b) ii
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